Question

respectively.
Find the zeros of the quadratic polynomial x^2 -13x + 36.​

Answer 1

${x}^{2} - 13x + 36 = 0 \\ {x}^{2} - 9x - 4x + 36 = 0 \\ x(x - 9) - 4(x - 9) = 0 \\( x - 9)(x - 4) = 0 \\ x - 9 = 0 \: or \: x - 4 = 0 \\ x = 9 \: or \: x = 4$

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Answer 2

Answer:

x=4,9

Step-by-step explanation:

x2 - 13x+26 = x2 - 9x- 4x +36

x(x-9)-4(x-9)= (x-9)(x-4)